Originally Posted by

**Swlabr** Chauchy's Theorem is used to prove that finite groups with elements of order a power of $\displaystyle p$ are of order $\displaystyle p^n$. However, in this case the elements are of order $\displaystyle p$. This is a simpler case, and we can use the formula,

$\displaystyle |HK|=|H|.|K|/|H \cap K|$

That is to say, if we look at the problem in my proof each element has order $\displaystyle p$. As each element has order $\displaystyle p$ it will generate $\displaystyle p-2$ other elements of order $\displaystyle p$ and so if we take a minimum generating set, $\displaystyle P_1, P_2, \ldots, P_n$, the intersection of each pair is trivial, $\displaystyle G=P_1P_2 \ldots P_n$ and the formula can be applied. Thus,

$\displaystyle |G|=|P_1|.|P_2| \ldots |P_n| = p^n \neq pq$

and we have the contradiction needed to salvage my earlier proof.

I hope that makes sense...

It should be noted that that formula has an elementary proof, and so I believe that my proof will count as "elementary". I hope.